3.97 \(\int \frac {x (a+b \tanh ^{-1}(c x))^2}{d+c d x} \, dx\)

Optimal. Leaf size=172 \[ -\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^2 d} \]

[Out]

(a+b*arctanh(c*x))^2/c^2/d+x*(a+b*arctanh(c*x))^2/c/d-2*b*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^2/d+(a+b*arctanh
(c*x))^2*ln(2/(c*x+1))/c^2/d-b^2*polylog(2,1-2/(-c*x+1))/c^2/d-b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/c^2
/d-1/2*b^2*polylog(3,1-2/(c*x+1))/c^2/d

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Rubi [A]  time = 0.31, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {5930, 5910, 5984, 5918, 2402, 2315, 5948, 6056, 6610} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^2 d}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 c^2 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c d} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcTanh[c*x])^2)/(d + c*d*x),x]

[Out]

(a + b*ArcTanh[c*x])^2/(c^2*d) + (x*(a + b*ArcTanh[c*x])^2)/(c*d) - (2*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)]
)/(c^2*d) + ((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/(c^2*d) - (b^2*PolyLog[2, 1 - 2/(1 - c*x)])/(c^2*d) - (b
*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^2*d) - (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*c^2*d)

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5918

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTanh[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 - c^2
*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 5930

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f)/e, Int[((f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p)/(
d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && GtQ[m, 0]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u])
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2
/(1 + c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx &=-\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d+c d x} \, dx}{c}+\frac {\int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^2 d}-\frac {(2 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{d}-\frac {(2 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c d}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^2 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^2 d}-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c d}+\frac {b^2 \int \frac {\text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c d}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^2 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^2 d}+\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c d}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^2 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^2 d}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c^2 d}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c d}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^2 d}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^2 d}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^2 d}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^2 d}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 c^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 140, normalized size = 0.81 \[ \frac {2 a \left (a c x-a \log (c x+1)+b \log \left (1-c^2 x^2\right )\right )-2 b \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right ) \left (a+b \tanh ^{-1}(c x)-b\right )+4 b \tanh ^{-1}(c x) \left ((a-b) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+a c x\right )-b^2 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )+2 b^2 \tanh ^{-1}(c x)^2 \left (c x+\log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-1\right )}{2 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(a + b*ArcTanh[c*x])^2)/(d + c*d*x),x]

[Out]

(2*b^2*ArcTanh[c*x]^2*(-1 + c*x + Log[1 + E^(-2*ArcTanh[c*x])]) + 4*b*ArcTanh[c*x]*(a*c*x + (a - b)*Log[1 + E^
(-2*ArcTanh[c*x])]) + 2*a*(a*c*x - a*Log[1 + c*x] + b*Log[1 - c^2*x^2]) - 2*b*(a - b + b*ArcTanh[c*x])*PolyLog
[2, -E^(-2*ArcTanh[c*x])] - b^2*PolyLog[3, -E^(-2*ArcTanh[c*x])])/(2*c^2*d)

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x \operatorname {artanh}\left (c x\right ) + a^{2} x}{c d x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))^2/(c*d*x+d),x, algorithm="fricas")

[Out]

integral((b^2*x*arctanh(c*x)^2 + 2*a*b*x*arctanh(c*x) + a^2*x)/(c*d*x + d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x}{c d x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))^2/(c*d*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2*x/(c*d*x + d), x)

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maple [C]  time = 0.57, size = 5361, normalized size = 31.17 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x))^2/(c*d*x+d),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} {\left (\frac {x}{c d} - \frac {\log \left (c x + 1\right )}{c^{2} d}\right )} + \frac {{\left (b^{2} c x - b^{2} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{4 \, c^{2} d} - \int -\frac {{\left (b^{2} c^{2} x^{2} - b^{2} c x\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c^{2} x^{2} - a b c x\right )} \log \left (c x + 1\right ) - 2 \, {\left ({\left (2 \, a b c^{2} + b^{2} c^{2}\right )} x^{2} - {\left (2 \, a b c - b^{2} c\right )} x + {\left (b^{2} c^{2} x^{2} - 2 \, b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c^{3} d x^{2} - c d\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))^2/(c*d*x+d),x, algorithm="maxima")

[Out]

a^2*(x/(c*d) - log(c*x + 1)/(c^2*d)) + 1/4*(b^2*c*x - b^2*log(c*x + 1))*log(-c*x + 1)^2/(c^2*d) - integrate(-1
/4*((b^2*c^2*x^2 - b^2*c*x)*log(c*x + 1)^2 + 4*(a*b*c^2*x^2 - a*b*c*x)*log(c*x + 1) - 2*((2*a*b*c^2 + b^2*c^2)
*x^2 - (2*a*b*c - b^2*c)*x + (b^2*c^2*x^2 - 2*b^2*c*x - b^2)*log(c*x + 1))*log(-c*x + 1))/(c^3*d*x^2 - c*d), x
)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{d+c\,d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*atanh(c*x))^2)/(d + c*d*x),x)

[Out]

int((x*(a + b*atanh(c*x))^2)/(d + c*d*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x}{c x + 1}\, dx + \int \frac {b^{2} x \operatorname {atanh}^{2}{\left (c x \right )}}{c x + 1}\, dx + \int \frac {2 a b x \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atanh(c*x))**2/(c*d*x+d),x)

[Out]

(Integral(a**2*x/(c*x + 1), x) + Integral(b**2*x*atanh(c*x)**2/(c*x + 1), x) + Integral(2*a*b*x*atanh(c*x)/(c*
x + 1), x))/d

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